Question

Find the right hand derivative and the left hand derivative of $\left \begin{array}{ll} f(x)= x -1 , & \quad x \ \textless \ 2 \\ \hspace{0.75cm}= 2x-3, & \quad x \geq 2 \end{array}$, at the point x = 2 and hence show that f(x) is not differentiable at x = 2.

Answer 1

Answer:

The right hand derivative is 2 and the left hand derivative is 1.

Step-by-step explanation:

right hand derivative:

$\lim_{h\to 0^+}\frac{f(2+h)-f(2)}{h} =\lim_{h\to 0^+} \frac{2(2+h)-3-1}{h} = \lim_{h\to 0^+}\frac{2h}{h} = 2$

left hand derivative:

$\lim_{h\to 0^-}\frac{f(2+h)-f(2)}{h} =\lim_{h\to 0^-} \frac{(2+h)-1-1}{h} = \lim_{h\to 0^+}\frac{h}{h} = 1$

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

GIVEN

$\sf{f(x) = \begin{cases} \sf{x} - 1 & \text{x < 2} \\ \sf{2x - 3} & \sf{x } \geqslant 2\end{cases}}$

TO DETERMINE

• Right hand derivative at x = 2

• Left hand derivative at x = 2

• TO show f(x) is not differentiable at x = 2

CALCULATION

Here

$\sf{f(x) = \begin{cases} \sf{x} - 1 & \text{x < 2} \\ \sf{2x - 3} & \sf{x } \geqslant 2\end{cases}}$

$\sf{So \: \: \: \: f(2) = 4 - 3 = 1}$

Now

Right hand derivative at x = 2

$= \displaystyle \sf{\lim_{x \to 2 {+}} \: \frac{f(x)-f(2)}{x-2}}$

$= \displaystyle \sf{\lim_{x \to 2 {+}} \: \frac{2x - 3-1}{x-2}}$

$= \displaystyle \sf{\lim_{x \to 2 {+}} \: \frac{2x - 4}{x-2}}$

$= \displaystyle \sf{\lim_{x \to 2 {+}} \: \frac{2(x - 2)}{x-2}}$

$= \displaystyle 2 \: \: \: \sf{\lim_{x \to 2 {+}} \: \frac{(x - 2)}{x-2}}$

$= \displaystyle \sf{2}$

Left hand derivative at x = 2

$= \displaystyle \sf{\lim_{x \to 2 { - }} \: \frac{f(x)-f(2)}{x-2}}$

$= \displaystyle \sf{\lim_{x \to 2 { - }} \: \frac{x - 1 - 1}{x-2}}$

$= \displaystyle \sf{\lim_{x \to 2 { - }} \: \frac{x - 2}{x-2}}$

$= \displaystyle \sf{1}$

Hence Right hand derivative at x = 2 is 2 and

Left hand derivative at x = 2 is 1

Hence at x = 2

Right hand derivative # Left hand derivative

Hence f(x) is not differentiable at x = 2

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